Answer to Question #311112 in Molecular Physics | Thermodynamics for Samsam

Question

A closed tank contains 0.5m of mercury density 13600 kg/m3 , 2m of water of density 1000 kg/m3 and 3m of oil of density 900 kg/m3 and there is an air space above the oil. The gauge pressure at the bottom of the tank is 200 kPa.

(i) Illustrate the stratification of the fluids under gravity with aid of diagram.

(ii) Determine the gauge pressure at the top of the tank.

Answer 1

The gauge pressure at the top of the tank

$$p_1=p_2-(g\rho_1h_1+g\rho_2h_2+g\rho_3h_3)\\ =2*10^5-9.8(13600*0.5+1000*2+900*3)\\ =87300\:\rm Pa$$